Re: Generators of GxH

(a, b) generates GxH is given. You don't assume premises - you generally "assume" things you want to prove false by deriving a contradiction from them. The way I was going was beginning a proof by contradiction, although it could (should) have been done directly.

Rather than work it out, the REASON it's true is:
(a,b) is a generator, so (a^n,b^n) hits every element of GxH, and so it hits (g,1) for every g in G. Thus for every g in G, there's some m that makes (a^m, b^m) = (g, 1), and, in particular, makes a^m = g.